Question: Simplify and expand the following expression: $ \dfrac{-7}{z + 2}+\dfrac{8}{z - 2} $
Explanation: In order to add expressions, they must have a common denominator. Get both fractions over a common denominator of $(z + 2)(z - 2)$ Multiply the first term by $\dfrac{z - 2}{z - 2}$ $ \begin{align*} \dfrac{-7}{z + 2} \times \dfrac{z - 2}{z - 2} & = \dfrac{(-7)(z - 2)}{(z + 2)(z - 2)} \\ & = \dfrac{-7z + 14}{(z + 2)(z - 2)}\end{align*} $ Multiply the second term by $\dfrac{z + 2}{z + 2}$ $ \begin{align*} \dfrac{8}{z - 2} \times \dfrac{z + 2}{z + 2} & = \dfrac{(8)(z + 2)}{(z - 2)(z + 2)} \\ & = \dfrac{8z + 16}{(z - 2)(z + 2)}\end{align*} $ Now we have: $ = \dfrac{-7z + 14}{(z + 2)(z - 2)} + \dfrac{8z + 16}{(z - 2)(z + 2)} $ Now both terms have a common denominator we can simply add the numerators: $ = \dfrac{-7z + 14 + 8z + 16}{(z + 2)(z - 2)} $ $ = \dfrac{z + 30}{(z + 2)(z - 2)}$ Expand the denominator: $ = \dfrac{z + 30}{z^2 - 4}$